Is There an Alternative Method for Proving a Limit Involving e and ln?

[Tomath]

Homework Statement

Show that lim (x,y) → (a,0) e^(x ln y) = 0 \foralla > 0

Homework Equations

The Attempt at a Solution

I've tried looking at lim (x,y) → (a,0) x ln y seperately.
lim(x,y) → (a,0) x ln y = lim(x,y) -> (a,0) x * lim(x,y) → (a,0) ln y
= a * lim(x,y) → (a,0) ln y

Now lim(x,y) → (a,0) ln y is -∞. So we get lim(x,y) →(a,0) x ln y = -∞. Now lim z → -∞ e^z = 0.

The problem here is that we haven't discussed limits involving infinity in class and I'm pretty sure I'm not allowed to use it. My question therefore is: is there any other way to show that lim (x,y) → (a,0) e^(x ln y) = 0?

 

[Dick]

Homework Statement

Show that lim (x,y) → (a,0) e^(x ln y) = 0 \foralla > 0

Homework Equations

The Attempt at a Solution

I've tried looking at lim (x,y) → (a,0) x ln y seperately.
lim(x,y) → (a,0) x ln y = lim(x,y) -> (a,0) x * lim(x,y) → (a,0) ln y
= a * lim(x,y) → (a,0) ln y

Now lim(x,y) → (a,0) ln y is -∞. So we get lim(x,y) →(a,0) x ln y = -∞. Now lim z → -∞ e^z = 0.

The problem here is that we haven't discussed limits involving infinity in class and I'm pretty sure I'm not allowed to use it. My question therefore is: is there any other way to show that lim (x,y) → (a,0) e^(x ln y) = 0?

No, I think that's the right way to do it.

 

[Tomath]

That's what I was afraid off :(. How would I prove that lim (x,y) → (a,0) e^(x ln y) = 0 and lim z → -∞ e^z = 0? I am familar with the \delta and \epsilon method but I am not sure how to include ∞ in this method.

 

[Dick]

That's what I was afraid off :(. How would I prove that lim (x,y) → (a,0) e^(x ln y) = 0 and lim z → -∞ e^z = 0? I am familar with the \delta and \epsilon method but I am not sure how to include ∞ in this method.

lim z → -∞ e^z = 0 just means that for any ε>0 you can find a N such that e^z<ε if z<N. Try picking N=log(ε).